Please solve for Part B The diagram shows a simplified household circuit. Resist

Question

Please solve for Part B

The diagram shows a simplified household circuit. Resistor R1 = 209.0 ohm represents a lightbulb; resistor R2 = 12.00 ohm represents a hair dryer. The resistors r = 0.4000 ohm (each) represent the resistance of the wiring in the walls. Assume that the generator supplies a constant 120.0 V rms. If the lightbulb is on and the hair dryer is off, find the rms voltage across the lightbulb and the power dissipated by the lightbulb. (b) If both the lightbulb and the hair dryer are on, find the rms voltage across the lightbulb, the power dissipated by the lightbulb, and the rms voltage between point A and ground, (c) Explain why lights sometimes dim when an appliance is turned on. (d) Explain why the neutral and ground wires in a junction box are not at the same potential even though they arc both grounded. Vrms = V, and P = W V1 = V, P1 = W, and VA to ground = V

Explanation / Answer

considering both are on

(2r+R2) is parallel to R1

so net is R1*(2r+R2)/{R1+R2+2r} = 12.06

now net resistance of circuit is 12.06+2r = 12.86

I(rms) = Vrms/ R(net) = 120/12.86 = 9.33 A

let I1 is rms current thorugh R1 and I2 is through R2

I(rms)= I1+I2

I1/I2 = (R2+2r)/R1 = 12.8/209 = 0.0612

I2 = 16.34*I1

I1+ 16.34*I1 = I(rms)

I1 = I(rms)/17.34 = 0.538A

so voltage across R1 or light bulb = R1*I1= 0.538*209=112.45 V

power dissipated by light bulb = V*I1= 112.45*0.538 = 60.4981 w

voltage between point A and ground is I(rms)*r = 9.33*0.4 =3.731 V

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---